# 高等数学——求解不定积分经典换元法

## 第一类换元法

$d[F(u)]=d[F(\phi(x)]=f[\phi(x)]\phi'(x)dx$

$\int f[\phi(x)]\phi'(x)dx = F[\phi(x)] + C = [\int f(u)du]_{u=\phi(x)}$

$\int g(x)dx = \int f[\phi(x)]\phi'(x) dx = [\int f(u)du]_{u=\phi(x)}$

$\int \frac{1}{3 + 2x}dx$

\begin{aligned} \int \frac{1}{3 + 2x}dx &= \int \frac{1}{2}\cdot \frac{1}{3+2x} d(3 + 2x) \\ &= \int \frac{1}{2} \cdot \frac{1}{u} du = \frac{1}{2} \ln|u| + C \\ &= \frac{1}{2} \ln|3 + 2x| + C \end{aligned}

$\int \cos^2xdx$

\begin{aligned} \int \cos^2x dx = \int \frac{1}{2}(1 + \cos 2x) dx \end{aligned}

\begin{aligned} \int \cos^2x dx &= \frac{1}{2} (\frac{1}{2} \int du + \int \frac{1}{2}\cos u du) \\ &= \frac{1}{2} \int dx + \frac{1}{4} \int \cos u du \\ &= \frac{x}{2} + \frac{\sin 2x}{4} + C \end{aligned}

## 第二类换元法

$\int f(x)dx = \int f[\phi(t)]\phi'(t)dt$

$\int f(x) dx = [\int f[\phi(t)]\phi'(t)dt]_{t = \phi^{-1}(x)}$

\begin{aligned} F'(x) &= \frac{d\Phi}{dt}\cdot \frac{dt}{dx} = f[\phi(t)]\phi'(t)\cdot \frac{1}{\phi'(t)} \\ &= f[\phi(t)] = f(x) \end{aligned}

\begin{aligned} \int \sqrt{a^2 - x^2}dx = \int a\cos t \cdot a\cos t dt = a^2 \int cos^2 tdt \end{aligned}

$\int cos^2 tdt$ 其实就是我们上面讲的第二个例子，我们之前计算得到过答案：$\int \cos^2x dx = \frac{x}{2} + \frac{\sin 2x}{4} + C$，我们代入原式，可以得到：

\begin{aligned} \int \sqrt{a^2 - x^2}dx &= a^2(\frac{t}{2} + \frac{\sin 2t}{4}) + C \\ &= \frac{a^2}{2}t + \frac{a^2}{2}\sin t \cos t + C \end{aligned}

$\int \sqrt{a^2 - x^2}dx = \frac{a^2}{2}\arcsin \frac{x}{a} + \frac{1}{2}x \sqrt{a^2 - x^2} + C$

posted @ 2020-03-27 08:40  TechFlow2019  阅读(...)  评论(...编辑  收藏